Writing exponential functions from graphs

- [Instructor] The graphs of the linear function f of x is equal to mx plus b and the exponential function g of x is equal to a times r to the x where r is greater than zero pass through the points negative one comma nine, so this is negative one comma nine right over here, and one comma one. Both graphs are given below. So this very clearly is the linear function. It is a line right over here. And this right over here is the exponential function, so right over here. And given the fact that this exponential function keeps decreasing as x gets larger and larger and larger is a pretty good hint that our r right over here, they tell us that r is greater than zero, but it's a pretty good hint that r is going to be between zero and one. The fact that g of x keeps approaching, it's getting closer and closer and closer to zero as x increases. But let's use the data they're giving us, the two points of intersection, to figure out what the equations of these two functions are. So first we can tackle the linear function. So f of x is equal to mx plus b. So they give us two points. We could use those points first to figure out the slope. So our m, our m right over here, that's our slope. That's our change in y over change in x. The rate and change of the vertical axis with respect to the horizontal axis. So let's see, between those two points, what is our change in x? Our change in x, we're going from x equals negative one to x equals one. X equals one. So we could think of it as we're finishing at one. We started at negative one. So one minus negative one, our change in x is two. We see that right over there. And what about our change in y? Well, we start at nine. Let me do this in maybe another color here. We start at nine, and we end up at one. So we end up at one. We started at nine. One minus nine is negative eight. And just to be clear, when x is one, y is one. When x is negative one, y is nine. Another way to think about it, the way I drew it right over here, we're finishing at x equals one, y equals one. We started at x equals negative one, y equals nine, and so we just took the differences. We get negative eight over two, which is equal to negative four. Which is equal to negative four. And so now we can write that f of x is equal to negative four, negative four, that's our slope, times x. Negative four times x plus b. And you can see that slope right over here. Every time you increase your... Every time you increase your x by one. I gotta be careful here, I got a little bit. Every time you increase your x by one, you're decreasing your y. And here on the x axis, we're marking off every half. So every time you increase your x by one, you are decreasing your y. You are decreasing your y by four there, so that makes sense that the slope is negative four. So now let's think about what b is. So to figure out b, we could use either one of these points to figure out, given an x, what f of x is, and then we can solve for by. Let's try one, 'cause one is a nice, simple number. So we could write f of one, which would be negative four times one plus b. And they tell us that f of one is one, is equal to one. And so this part right over here, we could write that as negative four plus b is equal to one, and then we could add four to both sides of this equation, and then we get b is equal to five. So we get f of x is equal to negative 4x plus five. Now, does that make sense that the y-intercept here is five? Well, you see that right over here. By inspection, you could have guessed, actually, that the y-intercept here is five, but now we've solved it. Maybe this was 5.00001 or something, but now we know for sure it's negative 4x plus five. Or another way you could've said it, if the slope is negative four, if this right over here is nine, you increase one in the x direction, you're gonna decrease four in the y direction, and that will get you to y is equal to five, so that is the y-intercept. But either way, we have figured out the linear function. Now let's figure out the exponential function. So here we could just use the two points to figure out these two unknowns. So, for example, let's try this first point. So g of negative one, which if we look at this right over here, would be a times r to the negative one. They tell us that g of negative one is going to be equal to nine. G of negative one is equal to nine. And so we could write this a times r to the negative one. That's the same thing as a over r is equal to nine, or we could multiply both sides by r, and we could say a is equal to 9r. Now let's use this other point. This other point, they tell us. They tell us that g of one, which would be the same thing as a times r to the first power or just a times r, that that is equal to one, or a times r is equal to one. So how can we use this information right here, a is equal to 9r and a times r is equal to one, to solve for a and r? Well, I have a little system here. It is a non-linear system, but it's a pretty simple one. We could just take this a and substitute it in right over here for a, and so we would get 9r for a. This first constraint tells us a must be equal to 9r. So 9r, instead of writing an a here, I'll write 9r, times r is equal to one. Or we could write, let me scroll down a little bit. We could write 9r squared is equal to one. Divide both sides by nine. R squared is equal to one over nine. And now to figure out r, you might wanna take the positive and negative square root of both sides, but they tell us that r is greater than zero, so we can just take the principal root of both sides and we get r is equal to 1/3. And then we could substitute this back into either one of these other two to figure out what a is. We know that a is equal to nine times r. So nine times 1/3, a is equal to three. So our exponential function could be written as g of x is equal to a, which is three, times r, which is 1/3, 1/3 to the x power.